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A piece of wood from the ruins of an ancient building was found to have a ${ }^{14} \mathrm{C}$ activity of 12 disintegrations per minute per gram of its carbon content. The ${ }^{14} \mathrm{C}$ activity of the living wood is 16 disintergrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given half-life of ${ }^{14} \mathrm{C}$ is $5760 \mathrm{yr}$.
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Verified Answer
Rate of disintegration in old wood sample of $\mathrm{C}-14$ radioactive atom is 12 atom per min per gram.
As, $\mathrm{R}=12 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, \mathrm{R}_0=16 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}$, $\mathrm{T}_{1 / 2}=5760 \mathrm{yr}$
Let $t$ be the span of the tree.
According to radioactive decay law,
$$
\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}
$$
or $\quad \mathrm{R}=\mathrm{R}_0 \mathrm{e}^{-\lambda \mathrm{t}}$ or $\frac{\mathrm{R}}{\mathrm{R}_0}=\mathrm{e}^{-\lambda \mathrm{t}}$ or $\mathrm{e}^{\lambda t}=\frac{\mathrm{R}_0}{\mathrm{R}}$
Taking $\log$ on both the sides
$$
\begin{aligned}
&\lambda t \log _{\mathrm{e}} \mathrm{e}=\log _{\mathrm{e}} \frac{\mathrm{R}_0}{\mathrm{R}} \Rightarrow \lambda \mathrm{t}=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\
&\mathrm{t}=\frac{2.303 \times \log _{10}\left[\frac{4}{3}\right]}{\lambda} \text { (half life) } \\
&\mathrm{t}=\frac{2.303(\log 4-\log 3)}{\lambda}
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad\left(\because \lambda=\frac{0.6931}{\mathrm{~T}_{1 / 2}}\right) \\
&=2391.20 \text { years. }
\end{aligned}
$$
As, $\mathrm{R}=12 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}, \mathrm{R}_0=16 \mathrm{dis} / \mathrm{min}$ per $\mathrm{g}$, $\mathrm{T}_{1 / 2}=5760 \mathrm{yr}$
Let $t$ be the span of the tree.
According to radioactive decay law,
$$
\mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}}
$$
or $\quad \mathrm{R}=\mathrm{R}_0 \mathrm{e}^{-\lambda \mathrm{t}}$ or $\frac{\mathrm{R}}{\mathrm{R}_0}=\mathrm{e}^{-\lambda \mathrm{t}}$ or $\mathrm{e}^{\lambda t}=\frac{\mathrm{R}_0}{\mathrm{R}}$
Taking $\log$ on both the sides
$$
\begin{aligned}
&\lambda t \log _{\mathrm{e}} \mathrm{e}=\log _{\mathrm{e}} \frac{\mathrm{R}_0}{\mathrm{R}} \Rightarrow \lambda \mathrm{t}=\left(\log _{10} \frac{16}{12}\right) \times 2.303 \\
&\mathrm{t}=\frac{2.303 \times \log _{10}\left[\frac{4}{3}\right]}{\lambda} \text { (half life) } \\
&\mathrm{t}=\frac{2.303(\log 4-\log 3)}{\lambda}
\end{aligned}
$$
$$
\begin{aligned}
&=\frac{2.303(0.6020-4.771) \times 5760}{0.6931} \quad\left(\because \lambda=\frac{0.6931}{\mathrm{~T}_{1 / 2}}\right) \\
&=2391.20 \text { years. }
\end{aligned}
$$
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